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Use the Chain Rule combined with other differentiation rules to find the derivative of the following functions. $$y=\left(\frac{e^{x}}{x+1}\right)^{8}$$

Short Answer

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Question: Find the derivative of the function $$y=\left(\frac{e^{x}}{x+1}\right)^{8}$$. Answer: The derivative of the function $$y=\left(\frac{e^{x}}{x+1}\right)^{8}$$ is $$\frac{dy}{dx} = 8\left(\frac{e^x}{x+1}\right)^7\cdot\frac{e^x(x)}{(x+1)^2}$$.

Step by step solution

01

Identify the inner and outer functions

In this case, we have an outer function $$y=u^8$$, where $$u=\frac{e^x}{x+1}$$. The inner function is thus $$u=\frac{e^x}{x+1}$$.
02

Differentiate the outer function with respect to u

Using the Power Rule, differentiate $$y=u^8$$ with respect to $$u$$: $$\frac{dy}{du} = 8u^7$$.
03

Differentiate the inner function with respect to x

Now, we need to differentiate the inner function $$u=\frac{e^x}{x+1}$$ with respect to $$x$$. This requires using the Quotient Rule, which states that if $$h(x)=\frac{f(x)}{g(x)}$$, then $$h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$$. Applying the Quotient Rule to $$u=\frac{e^x}{x+1}$$, we set $$f(x)=e^x$$ and $$g(x)=x+1$$. Differentiating with respect to x: $$f'(x)=\frac{d}{dx}(e^x)=e^x$$ (using the Exponential Rule) and $$g'(x)=\frac{d}{dx}(x+1)=1$$ (using the Power Rule). Then, we have $$\frac{du}{dx}=\frac{e^x(x+1)-e^x(1)}{(x+1)^2} = \frac{e^x(x)}{(x+1)^2}$$.
04

Apply the Chain Rule

Now, we can apply the Chain Rule, which states that if $$y=f(u)$$ and $$u=g(x)$$, then $$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$$. We have $$\frac{dy}{du}=8u^7$$ and $$\frac{du}{dx}=\frac{e^x(x)}{(x+1)^2}$$, so the derivative of $$y$$ with respect to $$x$$ is: $$\frac{dy}{dx}=8u^7\cdot\frac{e^x(x)}{(x+1)^2}$$.
05

Substitute the inner function back

Finally, substitute $$u=\frac{e^x}{x+1}$$ back into the expression of the derivative: $$\frac{dy}{dx} = 8\left(\frac{e^x}{x+1}\right)^7\cdot\frac{e^x(x)}{(x+1)^2}$$. Thus, the derivative of the function $$y=\left(\frac{e^{x}}{x+1}\right)^{8}$$ is $$\frac{dy}{dx} = 8\left(\frac{e^x}{x+1}\right)^7\cdot\frac{e^x(x)}{(x+1)^2}$$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The power rule is a basic technique used for differentiating functions that are raised to a power. In its simplest form, the power rule states that for any real number n, if y = x^n, then the derivative of y with respect to x is ny^{n-1}. Stated mathematically, this is:
\[ \frac{d}{dx}(x^n) = nx^{n-1} \].
This rule greatly simplifies the process of differentiation because it allows us to quickly find the derivative without resorting to the limit definition of a derivative each time. In the exercise provided, the power rule was applied in Step 2 after recognizing that the outer function can be expressed as u raised to the eighth power. By taking the derivative with respect to the inner function u, we get 8u^7, which follows directly from the power rule.
Quotient Rule
The quotient rule is essential for finding the derivative of a function that is the quotient of two differentiable functions. This rule states that if you have a function h(x) = f(x) / g(x), where both f(x) and g(x) are differentiable, the derivative h'(x) is given by:
\[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \].
This formula may appear complex, but it essentially says that the derivative of a quotient is the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all over the square of the denominator. In the initial problem, to differentiate the inner function u, which is a quotient of an exponential function and a linear function, the quotient rule comes into play, as shown in Step 3. It is critical to accurately differentiate the numerator and the denominator separately before applying the rule itself.
Exponential Function Differentiation
Differentiating exponential functions requires understanding the unique property of the exponential function. The derivative of an exponential function is proportional to the function itself. For any base a, the derivative of a^x is a^x times the natural logarithm of the base a. The most common exponential function in calculus is e^x, which has a derivative that is simply e^x itself. Formally, if y = e^x, then:
\[ \frac{d}{dx}(e^x) = e^x \].
In our exercise case, this simple derivative is part of the inner function u as indicated in Step 3. Whenever differentiating an exponential term like e^x, remember this unique property. This rule is also known as the Exponential Rule and plays a key role in simplifying many calculus problems dealing with exponential growth or decay.

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Most popular questions from this chapter

Product Rule for three functions Assume that \(f, g,\) and \(h\) are differentiable at \(x\) a. Use the Product Rule (twice) to find a formula for \(\frac{d}{d x}(f(x) g(x) h(x))\) b. Use the formula in (a) to find \(\frac{d}{d x}\left(e^{2 x}(x-1)(x+3)\right)\)

a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}-2 x\right)^{2}=2\left(x^{2}+y^{2}\right)\) \(\left(x_{0}, y_{0}\right)=(2,2)\) (limaçon of Pascal)

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{x \rightarrow e} \frac{\ln x-1}{x-e}$$

The flow of a small stream is monitored for 90 days between May 1 and August 1. The total water that flows past a gauging station is given by $$V(t)=\left\\{\begin{array}{ll}\frac{4}{5} t^{2} & \text { if } 0 \leq t<45 \\\\-\frac{4}{5}\left(t^{2}-180 t+4050\right) & \text { if } 45 \leq t<90, \end{array}\right.$$ where \(V\) is measured in cubic feet and \(t\) is measured in days, with \(t=0\) corresponding to May 1. a. Graph the volume function. b. Find the flow rate function \(V^{\prime}(t)\) and graph it. What are the units of the flow rate? c. Describe the flow of the stream over the 3 -month period. Specifically, when is the flow rate a maximum?

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h}$$

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